Let's take as example frames the GCRF (inertial, Earth-centered, non rotating) and ITRF (non-inertial, Earth-centered, rotating).
Previous questions have discussed the conversion of position and velocity.
However, there is still one key component missing, which is the conversion of acceleration vectors between the two frames. I believe this is in fact key, since, for example, an accurate acceleration due to the Earth gravitational attraction can be calculated by using spherical harmonics, which will result in an acceleration in ITRF which needs to be converted back to GCRF.
I have checked Vallado's Fundamentals of Astrodynamics and Applications, section 3.4, where the following equation is presented (the original equation uses subindices Inertial and Rot instead of GCRF and ITRF respectively; I have replaced them here by GCRF and ITRF, as they are the examples of inertial and rotating frames that I am currently considering):
$$ \vec{a}_{GCRF} = \vec{a}_{ITRF} + \dot{\vec{\omega}}\ \times\ \vec{r}_{ITRF} + \vec{\omega}\ \times\ \left(\vec{\omega}\ \times\ \vec{r}_{ITRF} \right) + 2\ \vec{\omega}\ \times\ \vec{v}_{ITRF} $$
Where:
- $\vec{a}_{GCRF}$ is the acceleration in GCRF frame
- $\vec{a}_{ITRF}$ is the acceleration in ITRF frame
- $\vec{r}_{ITRF}$ is the position in ITRF frame
- $\vec{v}_{ITRF}$ is the velocity in ITRF frame
- $\vec{\omega}$ is the angular velocity vector of Earth, which as David Hammen explained in his answer here, can be quite difficult to calculate in a very exact manner, but approximate ways seem relatively straightforward
- $\dot{\vec{\omega}}$ would be, if I understand correctly, the time-derivative of said angular velocity vector, i.e., a measurement of the rate of change of the rotation of Earth
I thought that $\dot{\vec{\omega}}$ would be very close to 0, and therefore that term would be ignored. However, Vallado goes on after that equation to state:
Notice the rotating acceleration is augmented by several other parameters to account for the coordinate system rotation. The supplemental terms are usually identified with specific names. The first term ($\dot{\vec{\omega}}\ \times\ \vec{r}_{ITRF}$) is called the tangential acceleration because $\vec{\omega}$ changes. For the satellite problems, it’s zero only for a true circular orbit. Next are the centripetal acceleration, ($\vec{\omega}\ \times\ \left(\vec{\omega}\ \times\ \vec{r}_{ITRF} \right)$), and the Coriolis acceleration, ($2\ \vec{\omega}\ \times\ \vec{v}_{ITRF}$)
(Emphasis added by me; I have also modified the exact mathematic notation used in the quote to match the equation with modified notation I presented previously).
It seems then that such term would in fact not be 0 for all orbits that are not circular. But I am surprised by this, why would this term be 0 or different than 0 depending on the eccentricity of the orbit of the satellite? I thought it would be independent of the satellite, since it is a measurement of the rate of change of Earth's angular velocity.
To add to the confusion, I have found other sources, such as this thesis, which state basically that the acceleration can be directly converted between GCRF and ITRF frames by simply applying the same transformation that is applied to position vectors (see equation 2.73 at the end of page 30 of the linked document and the preceding paragraph).
So, which would be the correct procedure? If the procedure presented by Vallado is in fact the correct one (which, in principle, makes more sense to me), what is exactly the term involving $\dot{\vec{\omega}}\ \times\ \vec{r}_{ITRF}$?
